I am learning interesting topics in set theory class. I believe that there is a straight proof of Axiom of Choice II by Axiom of Choice I. If you don't know the axioms, see below: Axiom of Choice I states that for any relation R, there is a function [itex]F\subseteq R[/itex] with [itex]dom(F) = dom(R)[/itex]. Axiom of Choice II states that the Cartesian product of nonempty sets is always nonempty. That is, if H is a function with domain I and if [itex](\forall i \in I)H(i)\neq ∅[/itex], then there is a function f with domain I such that [itex](\forall i \in I)f(i) \subseteq H(i)[/itex]. The question is: Is there a way to prove Axiom of Choice I from Axiom of Choice II?